Thread: Eigenvalues and eigenvectors

Let A be a n x n matrix such that A^k = 0 (k some natural number).
Prove that A^n = 0.

Any ideas?
Thanks.

Let v be an eigenvector. Then Av=cv, where c is the corresponding eigenvalue. A^k v=c^k v=0. Then c must be 0. Therefore, there is only one eigenvalue. Now J=P^-1 A P, where J is the Jordan normal form of the matrix and P is a non singular matrix. You can easily check that J^n=P^-1 A^n P. Now, the Jordan normal form is either 0, in which case, Ju=P^-1 A P u=0, for any vector u, which implies that A is 0; or the Jordan normal form is a matrix with all entries 0, except above the diagonal where they are 1's (some authors prefer to put the 1's below the diagonal). Now you can also check that after n multiplications of any vector u with J^n, it will be 0, which implies that A^n is 0.

the fact that A^k = 0, means that the minimal polynomial for A, m(x), divides x^k (*), so is of the form x^t, for some t.

the fact that every root of the characteristic polynomial (eigenvalue) is also a root of the minimal polynomial (**), means that the characteristic polynomial of A is:

(x - 0)ⁿ = xⁿ.

since every matrix A satisfies its characteristic polynomial (Cayley-Hamilton), Aⁿ = 0.

a proof of (*):

suppose that p(A) = 0, for some polynomial p(x), and let m(x) be the minimal polynomial for A.

writing p(x) = q(x)m(x) + r(x), where either r(x) = 0, or deg(r) < deg(m), we have:

0 = p(A) = q(A)m(A) + r(A) = q(A)(0) + r(A) = r(A).

if deg(r) < deg(m), this contradicts the minimality of m, so we must have r(x) = 0 (the 0-polynomial), so that p(x) = q(x)m(x), so m(x) divides p(x).

a proof of (**):

suppose λ is an eigenvalue of A (and thus a root of det(xI - A)), so that for some eigenvector v: A(v) = λv, and let the minimal polynomial of A be m(x), so that m(A) = 0 (the 0-matrix).

then (m(A))(v) = (m(λ))v. but (m(A))v = 0v = 0, so that m(λ) must be 0, that is, λ is a root of m(x).

Is my solution correct, Deveno?

yes! the JNF of such a matrix must consist of 0's on the diagonal, and some number of 1's on the superdiagonal. so:

A = PNP^-1, where N is of nilpotency ≤ n. whence:

Aⁿ = PNⁿP^-1 = P(0)P^-1 = 0.

(here, my "N" is your "J"...i typically write J = D+N, where D is diagonal and N is nilpotent. the first part of your post shows that D = 0).

postscript: my answer was in no way intended to invalidate ModusPonens' excellent reply. i wanted to illustrate the richness of linear algebra: often, there are several ways to look at (and solve) a given problem.

It wasn't my opinion that you responded to invalidate my answer, of course!